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邝斌计算几何模版

二维平面

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// 计算几何模板
//二维平面
using namespace std;
const double eps = 1e-8;
const double inf = 1e20;
const double pi = acos(-1.0);
const int maxp = 1010;
//Compares a double to zero
int sgn(double x){
if(fabs(x) < eps)return 0;
if(x < 0)return -1;
else return 1;
}
//square of a double
inline double sqr(double x){return x*x;}

struct Point{
double x,y;
Point(){}
Point(double _x,double _y){
x = _x;
y = _y;
}
void input(){
scanf("%lf%lf",&x,&y);
}
void output(){
printf("%.2f-%.2f\n",x,y);
}
bool operator == (Point b)const{
return sgn(x-b.x) == 0 && sgn(y-b.y) == 0;
}
bool operator < (Point b)const{
return sgn(x-b.x)== 0-sgn(y-b.y)?0:x<b.x;
}
Point operator-(const Point &b)const{
return Point(x-b.x,y-b.y);
}
//叉积
double operator ^(const Point &b)const{
return x*b.y-y*b.x;
}
//点积
double operator *(const Point &b)const{
return x*b.x + y*b.y;
}
//返回长度
double len(){
return hypot(x,y);//库函数
}
//返回长度的平方
double len2(){
return x*x + y*y;
}
//返回两点的距离
double distance(Point p){
return hypot(x-p.x,y-p.y);
}
Point operator +(const Point &b)const{
return Point(x+b.x,y+b.y);
}
Point operator *(const double &k)const{
return Point(x*k,y*k);
}
Point operator /(const double &k)const{
return Point(x/k,y/k);
}
//计算 pa 和 pb 的夹角
//就是求这个点看 a,b 所成的夹角
//测试 LightOJ1203
double rad(Point a,Point b){
Point p = *this;
return fabs(atan2( fabs((a-p)^(b-p)),(a-p)*(b-p) ));
}
//化为长度为 r 的向量
Point trunc(double r){
double l = len();
if(!sgn(l))return *this;
r /= l;
return Point(x*r,y*r);
}
//逆时针旋转 90 度
Point rotleft(){
return Point(-y,x);
}
//顺时针旋转 90 度
Point rotright(){
return Point(y,-x);
}
//绕着 p 点逆时针旋转 angle
Point rotate(Point p,double angle){
Point v = (*this)-p;
double c = cos(angle), s = sin(angle);
return Point(p.x + v.x*c-v.y*s,p.y + v.x*s + v.y*c);
}
};
struct Line{
Point s,e;
Line(){}
Line(Point _s,Point _e){
s = _s;
e = _e;
}
bool operator ==(Line v){
return (s == v.s)&&(e == v.e);
}
//根据一个点和倾斜角 angle 确定直线,0<=angle<pi
Line(Point p,double angle){
s = p;
if(sgn(angle-pi/2) == 0){
e = (s + Point(0,1));
}
else{
e = (s + Point(1,tan(angle)));
}
}
//ax+by+c=0
Line(double a,double b,double c){
if(sgn(a) == 0){
s = Point(0,-c/b);
e = Point(1,-c/b);
}
else if(sgn(b) == 0){
s = Point(-c/a,0);
e = Point(-c/a,1);
}
else{
s = Point(0,-c/b);
e = Point(1,(-c-a)/b);
}
}
void input(){
s.input();
e.input();
}
void adjust(){
if(e < s){
swap(s,e);
}
}
//求线段长度
double length(){
return s.distance(e);
}
//返回直线倾斜角 0<=angle<pi
double angle(){
double k = atan2(e.y-s.y,e.x-s.x);
if(sgn(k) < 0)k += pi;
if(sgn(k-pi) == 0)k-= pi;
return k;
}
//点和直线关系
//1 在左侧
//2 在右侧
//3 在直线上
int relation(Point p){
int c = sgn((p-s)^(e-s));
if(c < 0)return 1;
else if(c > 0)return 2;
else return 3;
}
// 点在线段上的判断
bool pointonseg(Point p){
return sgn((p-s)^(e-s)) == 0 && sgn((p-s)*(p-e)) <= 0;
}
//两向量平行 (对应直线平行或重合)
bool parallel(Line v){
return sgn((e-s)^(v.e-v.s)) == 0;
}
//两线段相交判断
//2 规范相交
//1 非规范相交
//0 不相交
int segcrossseg(Line v){
int d1 = sgn((e-s)^(v.s-s));
int d2 = sgn((e-s)^(v.e-s));
int d3 = sgn((v.e-v.s)^(s-v.s));
int d4 = sgn((v.e-v.s)^(e-v.s));
if( (d1^d2)==-2 && (d3^d4)==-2 )return 2;
return (d1==0 && sgn((v.s-s)*(v.s-e))<=0) ||
(d2==0 && sgn((v.e-s)*(v.e-e))<=0) ||
(d3==0 && sgn((s-v.s)*(s-v.e))<=0) ||
(d4==0 && sgn((e-v.s)*(e-v.e))<=0);
}
//直线和线段相交判断
//-*this line -v seg
//2 规范相交
//1 非规范相交
//0 不相交
int linecrossseg(Line v){
int d1 = sgn((e-s)^(v.s-s));
int d2 = sgn((e-s)^(v.e-s));
if((d1^d2)==-2) return 2;
return (d1==0||d2==0);
}
//两直线关系
//0 平行
//1 重合
//2 相交
int linecrossline(Line v){
if((*this).parallel(v))
return v.relation(s)==3;
return 2;
}
//求两直线的交点
//要保证两直线不平行或重合
Point crosspoint(Line v){
double a1 = (v.e-v.s)^(s-v.s);
double a2 = (v.e-v.s)^(e-v.s);
return Point((s.x*a2-e.x*a1)/(a2-a1),(s.y*a2-e.y*a1)/(a2-a1
));
}
//点到直线的距离
double dispointtoline(Point p){
return fabs((p-s)^(e-s))/length();
}
//点到线段的距离
double dispointtoseg(Point p){
if(sgn((p-s)*(e-s))<0 || sgn((p-e)*(s-e))<0)
return min(p.distance(s),p.distance(e));
return dispointtoline(p);
}
//返回线段到线段的距离
//前提是两线段不相交,相交距离就是 0 了
double dissegtoseg(Line v){
return min(min(dispointtoseg(v.s),dispointtoseg(v.e)),min(v
.dispointtoseg(s),v.dispointtoseg(e)));
}
//返回点 p 在直线上的投影
Point lineprog(Point p){
return s + ( ((e-s)*((e-s)*(p-s)))/((e-s).len2()) );
}
//返回点 p 关于直线的对称点
Point symmetrypoint(Point p){
Point q = lineprog(p);
return Point(2*q.x-p.x,2*q.y-p.y);
}
};
//圆
struct circle{
Point p;//圆心
double r;//半径
circle(){}
circle(Point _p,double _r){
p = _p;
r = _r;
}
circle(double x,double y,double _r){
p = Point(x,y);
r = _r;
}
//三角形的外接圆
//需要 Point 的 + / rotate() 以及 Line 的 crosspoint()
//利用两条边的中垂线得到圆心
//测试:UVA12304
circle(Point a,Point b,Point c){
Line u = Line((a+b)/2,((a+b)/2)+((b-a).rotleft()));
Line v = Line((b+c)/2,((b+c)/2)+((c-b).rotleft()));
p = u.crosspoint(v);
r = p.distance(a);
}
//三角形的内切圆
//参数 bool t 没有作用,只是为了和上面外接圆函数区别
//测试:UVA12304
circle(Point a,Point b,Point c,bool t){
Line u,v;
double m = atan2(b.y-a.y,b.x-a.x), n = atan2(c.y-a.y,c.x-a.
x);
u.s = a;
u.e = u.s + Point(cos((n+m)/2),sin((n+m)/2));
v.s = b;
m = atan2(a.y-b.y,a.x-b.x) , n = atan2(c.y-b.y,c.x-b.x);
v.e = v.s + Point(cos((n+m)/2),sin((n+m)/2));
p = u.crosspoint(v);
r = Line(a,b).dispointtoseg(p);
}
//输入
void input(){
p.input();
scanf("%lf",&r);
}
//输出
void output(){
printf("%.2lf-%.2lf-%.2lf\n",p.x,p.y,r);
}
bool operator == (circle v){
return (p==v.p) && sgn(r-v.r)==0;
}
bool operator < (circle v)const{
return ((p<v.p)||((p==v.p)&&sgn(r-v.r)<0));
}
//面积
double area(){
return pi*r*r;
}
//周长
double circumference(){
return 2*pi*r;
}
//点和圆的关系
//0 圆外
//1 圆上
//2 圆内
int relation(Point b){
double dst = b.distance(p);
if(sgn(dst-r) < 0)return 2;
else if(sgn(dst-r)==0)return 1;
return 0;
}
//线段和圆的关系
//比较的是圆心到线段的距离和半径的关系
int relationseg(Line v){
double dst = v.dispointtoseg(p);
if(sgn(dst-r) < 0)return 2;
else if(sgn(dst-r) == 0)return 1;
return 0;
}
//直线和圆的关系
//比较的是圆心到直线的距离和半径的关系
int relationline(Line v){
double dst = v.dispointtoline(p);
if(sgn(dst-r) < 0)return 2;
else if(sgn(dst-r) == 0)return 1;
return 0;
}
//两圆的关系
//5 相离
//4 外切
//3 相交
//2 内切
//1 内含
//需要 Point 的 distance
//测试:UVA12304
int relationcircle(circle v){
double d = p.distance(v.p);
if(sgn(d-r-v.r) > 0)return 5;
if(sgn(d-r-v.r) == 0)return 4;
double l = fabs(r-v.r);
if(sgn(d-r-v.r)<0 && sgn(d-l)>0)return 3;
if(sgn(d-l)==0)return 2;
if(sgn(d-l)<0)return 1;
}
//求两个圆的交点,返回 0 表示没有交点,返回 1 是一个交点,2 是两个交点
//需要 relationcircle
//测试:UVA12304
int pointcrosscircle(circle v,Point &p1,Point &p2){
int rel = relationcircle(v);
if(rel == 1 || rel == 5)return 0;
double d = p.distance(v.p);
double l = (d*d+r*r-v.r*v.r)/(2*d);
double h = sqrt(r*r-l*l);
Point tmp = p + (v.p-p).trunc(l);
p1 = tmp + ((v.p-p).rotleft().trunc(h));
p2 = tmp + ((v.p-p).rotright().trunc(h));
if(rel == 2 || rel == 4)
return 1;
return 2;
}
//求直线和圆的交点,返回交点个数
int pointcrossline(Line v,Point &p1,Point &p2){
if(!(*this).relationline(v))return 0;
Point a = v.lineprog(p);
double d = v.dispointtoline(p);
d = sqrt(r*r-d*d);
if(sgn(d) == 0){
p1 = a;
p2 = a;
return 1;
}
p1 = a + (v.e-v.s).trunc(d);
p2 = a-(v.e-v.s).trunc(d);
return 2;
}
//得到过 a,b 两点,半径为 r1 的两个圆
int gercircle(Point a,Point b,double r1,circle &c1,circle &c2){
circle x(a,r1),y(b,r1);
int t = x.pointcrosscircle(y,c1.p,c2.p);
if(!t)return 0;
c1.r = c2.r = r;
return t;
}
//得到与直线 u 相切,过点 q, 半径为 r1 的圆
//测试:UVA12304
int getcircle(Line u,Point q,double r1,circle &c1,circle &c2){
double dis = u.dispointtoline(q);
if(sgn(dis-r1*2)>0)return 0;
if(sgn(dis) == 0){
c1.p = q + ((u.e-u.s).rotleft().trunc(r1));
c2.p = q + ((u.e-u.s).rotright().trunc(r1));
c1.r = c2.r = r1;
return 2;
}

Line u1 = Line((u.s + (u.e-u.s).rotleft().trunc(r1)),(u.e +
(u.e-u.s).rotleft().trunc(r1)));
Line u2 = Line((u.s + (u.e-u.s).rotright().trunc(r1)),(u.e
+ (u.e-u.s).rotright().trunc(r1)));
circle cc = circle(q,r1);
Point p1,p2;
if(!cc.pointcrossline(u1,p1,p2))cc.pointcrossline(u2,p1,p2)
;
c1 = circle(p1,r1);
if(p1 == p2){
c2 = c1;
return 1;
}
c2 = circle(p2,r1);
return 2;
}
//同时与直线 u,v 相切,半径为 r1 的圆
//测试:UVA12304
int getcircle(Line u,Line v,double r1,circle &c1,circle &c2,
circle &c3,circle &c4){
if(u.parallel(v))return 0;//两直线平行
Line u1 = Line(u.s + (u.e-u.s).rotleft().trunc(r1),u.e + (u
.e-u.s).rotleft().trunc(r1));
Line u2 = Line(u.s + (u.e-u.s).rotright().trunc(r1),u.e + (
u.e-u.s).rotright().trunc(r1));
Line v1 = Line(v.s + (v.e-v.s).rotleft().trunc(r1),v.e + (v
.e-v.s).rotleft().trunc(r1));
Line v2 = Line(v.s + (v.e-v.s).rotright().trunc(r1),v.e + (
v.e-v.s).rotright().trunc(r1));
c1.r = c2.r = c3.r = c4.r = r1;
c1.p = u1.crosspoint(v1);
c2.p = u1.crosspoint(v2);
c3.p = u2.crosspoint(v1);
c4.p = u2.crosspoint(v2);
return 4;
}
//同时与不相交圆 cx,cy 相切,半径为 r1 的圆
//测试:UVA12304
int getcircle(circle cx,circle cy,double r1,circle &c1,circle &
c2){
circle x(cx.p,r1+cx.r),y(cy.p,r1+cy.r);
int t = x.pointcrosscircle(y,c1.p,c2.p);
if(!t)return 0;
c1.r = c2.r = r1;
return t;
}

//过一点作圆的切线 (先判断点和圆的关系)
//测试:UVA12304
int tangentline(Point q,Line &u,Line &v){
int x = relation(q);
if(x == 2)return 0;
if(x == 1){
u = Line(q,q + (q-p).rotleft());
v = u;
return 1;
}
double d = p.distance(q);
double l = r*r/d;
double h = sqrt(r*r-l*l);
u = Line(q,p + ((q-p).trunc(l) + (q-p).rotleft().trunc(h)))
;
v = Line(q,p + ((q-p).trunc(l) + (q-p).rotright().trunc(h))
);
return 2;
}
//求两圆相交的面积
double areacircle(circle v){
int rel = relationcircle(v);
if(rel >= 4)return 0.0;
if(rel <= 2)return min(area(),v.area());
double d = p.distance(v.p);
double hf = (r+v.r+d)/2.0;
double ss = 2*sqrt(hf*(hf-r)*(hf-v.r)*(hf-d));
double a1 = acos((r*r+d*d-v.r*v.r)/(2.0*r*d));
a1 = a1*r*r;
double a2 = acos((v.r*v.r+d*d-r*r)/(2.0*v.r*d));
a2 = a2*v.r*v.r;
return a1+a2-ss;
}
//求圆和三角形 pab 的相交面积
//测试:POJ3675 HDU3982 HDU2892
double areatriangle(Point a,Point b){
if(sgn((p-a)^(p-b)) == 0)return 0.0;
Point q[5];
int len = 0;
q[len++] = a;
Line l(a,b);
Point p1,p2;
if(pointcrossline(l,q[1],q[2])==2){
if(sgn((a-q[1])*(b-q[1]))<0)q[len++] = q[1];
if(sgn((a-q[2])*(b-q[2]))<0)q[len++] = q[2];
}
q[len++] = b;
if(len == 4 && sgn((q[0]-q[1])*(q[2]-q[1]))>0)swap(q[1],q
[2]);
double res = 0;
for(int i = 0;i < len-1;i++){
if(relation(q[i])==0||relation(q[i+1])==0){
double arg = p.rad(q[i],q[i+1]);
res += r*r*arg/2.0;
}
else{
res += fabs((q[i]-p)^(q[i+1]-p))/2.0;
}
}
return res;
}
};

struct polygon{
int n;
Point p[maxp];
Line l[maxp];
void input(int _n){
n = _n;
for(int i = 0;i < n;i++)
p[i].input();
}
void add(Point q){
p[n++] = q;
}
void getline(){
for(int i = 0;i < n;i++){
l[i] = Line(p[i],p[(i+1)%n]);
}
}
struct cmp{
Point p;
cmp(const Point &p0){p = p0;}
bool operator()(const Point &aa,const Point &bb){
Point a = aa, b = bb;
int d = sgn((a-p)^(b-p));
if(d == 0){
return sgn(a.distance(p)-b.distance(p)) < 0;
}
return d > 0;
}
};
//进行极角排序
//首先需要找到最左下角的点
//需要重载号好 Point 的 < 操作符 (min 函数要用)
void norm(){
Point mi = p[0];
for(int i = 1;i < n;i++)mi = min(mi,p[i]);
sort(p,p+n,cmp(mi));
}
//得到凸包
//得到的凸包里面的点编号是 0~n-1 的
//两种凸包的方法
//注意如果有影响,要特判下所有点共点,或者共线的特殊情况
//测试 LightOJ1203 LightOJ1239
void getconvex(polygon &convex){
sort(p,p+n);
convex.n = n;
for(int i = 0;i < min(n,2);i++){
convex.p[i] = p[i];
}
if(convex.n == 2 && (convex.p[0] == convex.p[1]))convex.n
--;//特判
if(n <= 2)return;
int &top = convex.n;
top = 1;
for(int i = 2;i < n;i++){
while(top && sgn((convex.p[top]-p[i])^(convex.p[top-1]-
p[i])) <= 0)
top--;
convex.p[++top] = p[i];
}
int temp = top;
convex.p[++top] = p[n-2];
for(int i = n-3;i >= 0;i--){
while(top != temp && sgn((convex.p[top]-p[i])^(convex.p
[top-1]-p[i])) <= 0)
top--;
convex.p[++top] = p[i];
}
if(convex.n == 2 && (convex.p[0] == convex.p[1]))convex.n
--;//特判
convex.norm();//原来得到的是顺时针的点,排序后逆时针
}
//得到凸包的另外一种方法
//测试 LightOJ1203 LightOJ1239
void Graham(polygon &convex){
norm();
int &top = convex.n;
top = 0;
if(n == 1){
top = 1;
convex.p[0] = p[0];
return;
}
if(n == 2){
top = 2;
convex.p[0] = p[0];
convex.p[1] = p[1];
if(convex.p[0] == convex.p[1])top--;
return;
}
convex.p[0] = p[0];
convex.p[1] = p[1];
top = 2;
for(int i = 2;i < n;i++){
while( top > 1 && sgn((convex.p[top-1]-convex.p[top-2])
^(p[i]-convex.p[top-2])) <= 0 )
top--;
convex.p[top++] = p[i];
}
if(convex.n == 2 && (convex.p[0] == convex.p[1]))convex.n
--;//特 判
}
//判断是不是凸的
bool isconvex(){
bool s[2];
memset(s,false,sizeof(s));
for(int i = 0;i < n;i++){
int j = (i+1)%n;
int k = (j+1)%n;
s[sgn((p[j]-p[i])^(p[k]-p[i]))+1] = true;
if(s[0] && s[2])return false;
}
return true;
}
//判断点和任意多边形的关系
// 3 点上
// 2 边上
// 1 内部
// 0 外部
int relationpoint(Point q){
for(int i = 0;i < n;i++){
if(p[i] == q)return 3;
}
getline();
for(int i = 0;i < n;i++){
if(l[i].pointonseg(q))return 2;
}
int cnt = 0;
for(int i = 0;i < n;i++){
int j = (i+1)%n;
int k = sgn((q-p[j])^(p[i]-p[j]));
int u = sgn(p[i].y-q.y);
int v = sgn(p[j].y-q.y);
if(k > 0 && u < 0 && v >= 0)cnt++;
if(k < 0 && v < 0 && u >= 0)cnt--;
}
return cnt != 0;
}
//直线 u 切割凸多边形左侧
//注意直线方向
//测试:HDU3982
void convexcut(Line u,polygon &po){
int &top = po.n;//注意引用
top = 0;
for(int i = 0;i < n;i++){
int d1 = sgn((u.e-u.s)^(p[i]-u.s));
int d2 = sgn((u.e-u.s)^(p[(i+1)%n]-u.s));
if(d1 >= 0)po.p[top++] = p[i];
if(d1*d2 < 0)po.p[top++] = u.crosspoint(Line(p[i],p[(i
+1)%n]));
}
}
//得到周长
//测试 LightOJ1239
double getcircumference(){
double sum = 0;
for(int i = 0;i < n;i++){
sum += p[i].distance(p[(i+1)%n]);
}
return sum;
}
//得到面积
double getarea(){
double sum = 0;
for(int i = 0;i < n;i++){
sum += (p[i]^p[(i+1)%n]);
}
return fabs(sum)/2;
}
//得到方向
// 1 表示逆时针,0 表示顺时针
bool getdir(){
double sum = 0;
for(int i = 0;i < n;i++)
sum += (p[i]^p[(i+1)%n]);
if(sgn(sum) > 0)return 1;
return 0;
}
//得到重心
Point getbarycentre(){
Point ret(0,0);
double area = 0;
for(int i = 1;i < n-1;i++){
double tmp = (p[i]-p[0])^(p[i+1]-p[0]);
if(sgn(tmp) == 0)continue;
area += tmp;
ret.x += (p[0].x+p[i].x+p[i+1].x)/3*tmp;
ret.y += (p[0].y+p[i].y+p[i+1].y)/3*tmp;
}
if(sgn(area)) ret = ret/area;
return ret;
}
//多边形和圆交的面积
//测试:POJ3675 HDU3982 HDU2892
double areacircle(circle c){
double ans = 0;
for(int i = 0;i < n;i++){
int j = (i+1)%n;
if(sgn( (p[j]-c.p)^(p[i]-c.p) ) >= 0)
ans += c.areatriangle(p[i],p[j]);
else ans-= c.areatriangle(p[i],p[j]);
}
return fabs(ans);
}
//多边形和圆关系
// 2 圆完全在多边形内
// 1 圆在多边形里面,碰到了多边形边界
// 0 其它
int relationcircle(circle c){
getline();
int x = 2;
if(relationpoint(c.p) != 1)return 0;//圆心不在内部
for(int i = 0;i < n;i++){
if(c.relationseg(l[i])==2)return 0;
if(c.relationseg(l[i])==1)x = 1;
}
return x;
}
};
//AB X AC
double cross(Point A,Point B,Point C){
return (B-A)^(C-A);
}
//AB*AC
double dot(Point A,Point B,Point C){
return (B-A)*(C-A);
}
//最小矩形面积覆盖
// A 必须是凸包 (而且是逆时针顺序)
// 测试 UVA 10173
double minRectangleCover(polygon A){
//要特判 A.n < 3 的情况
if(A.n < 3)return 0.0;
A.p[A.n] = A.p[0];
double ans =-1;
int r = 1, p = 1, q;
for(int i = 0;i < A.n;i++){
//卡出离边 A.p[i] - A.p[i+1] 最远的点
while( sgn( cross(A.p[i],A.p[i+1],A.p[r+1])-cross(A.p[i],
A.p[i+1],A.p[r]) ) >= 0 )
r = (r+1)%A.n;
//卡出 A.p[i] - A.p[i+1] 方向上正向 n 最远的点
while(sgn( dot(A.p[i],A.p[i+1],A.p[p+1])-dot(A.p[i],A.p[i
+1],A.p[p]) ) >= 0 )
p = (p+1)%A.n;
if(i == 0)q = p;
//卡出 A.p[i] - A.p[i+1] 方向上负向最远的点
while(sgn(dot(A.p[i],A.p[i+1],A.p[q+1])-dot(A.p[i],A.p[i
+1],A.p[q])) <= 0)
q = (q+1)%A.n;
double d = (A.p[i]-A.p[i+1]).len2();
double tmp = cross(A.p[i],A.p[i+1],A.p[r]) *
(dot(A.p[i],A.p[i+1],A.p[p])-dot(A.p[i],A.p[i+1],A.p[
q]))/d;
if(ans < 0 || ans > tmp)ans = tmp;
}
return ans;
}



//直线切凸多边形
//多边形是逆时针的,在 q1q2 的左侧
//测试:HDU3982
vector<Point> convexCut(const vector<Point> &ps,Point q1,Point q2){
vector<Point>qs;
int n = ps.size();
for(int i = 0;i < n;i++){
Point p1 = ps[i], p2 = ps[(i+1)%n];
int d1 = sgn((q2-q1)^(p1-q1)), d2 = sgn((q2-q1)^(p2-q1));
if(d1 >= 0)
qs.push_back(p1);
if(d1 * d2 < 0)
qs.push_back(Line(p1,p2).crosspoint(Line(q1,q2)));
}
return qs;
}
//半平面交
//测试 POJ3335 POJ1474 POJ1279
//***************************
struct halfplane:public Line{
double angle;
halfplane(){}
//表示向量 s->e 逆时针 (左侧) 的半平面
halfplane(Point _s,Point _e){
s = _s;
e = _e;
}
halfplane(Line v){
s = v.s;
e = v.e;
}
void calcangle(){
angle = atan2(e.y-s.y,e.x-s.x);
}
bool operator <(const halfplane &b)const{
return angle < b.angle;
}
};
struct halfplanes{
int n;
halfplane hp[maxp];
Point p[maxp];
int que[maxp];
int st,ed;
void push(halfplane tmp){
hp[n++] = tmp;
}
//去重
void unique(){
int m = 1;
for(int i = 1;i < n;i++){
if(sgn(hp[i].angle-hp[i-1].angle) != 0)
hp[m++] = hp[i];
else if(sgn( (hp[m-1].e-hp[m-1].s)^(hp[i].s-hp[m-1].s)
) > 0)
hp[m-1] = hp[i];
}
n = m;
}
bool halfplaneinsert(){
for(int i = 0;i < n;i++)hp[i].calcangle();
sort(hp,hp+n);
unique();
que[st=0] = 0;
que[ed=1] = 1;
p[1] = hp[0].crosspoint(hp[1]);
for(int i = 2;i < n;i++){
while(st<ed && sgn((hp[i].e-hp[i].s)^(p[ed]-hp[i].s))
<0)ed--;
while(st<ed && sgn((hp[i].e-hp[i].s)^(p[st+1]-hp[i].s))
<0)st++;
que[++ed] = i;
if(hp[i].parallel(hp[que[ed-1]]))return false;
p[ed]=hp[i].crosspoint(hp[que[ed-1]]);
}
while(st<ed && sgn((hp[que[st]].e-hp[que[st]].s)^(p[ed]-hp[
que[st]].s))<0)ed--;
while(st<ed && sgn((hp[que[ed]].e-hp[que[ed]].s)^(p[st+1]-
hp[que[ed]].s))<0)st++;
if(st+1>=ed)return false;
return true;
}
//得到最后半平面交得到的凸多边形
//需要先调用 halfplaneinsert() 且返回 true
void getconvex(polygon &con){
p[st] = hp[que[st]].crosspoint(hp[que[ed]]);
con.n = ed-st+1;
for(int j = st,i = 0;j <= ed;i++,j++)
con.p[i] = p[j];
}
};
//***************************

const int maxn = 1010;
struct circles{
circle c[maxn];
double ans[maxn];//ans[i] 表示被覆盖了 i 次的面积
double pre[maxn];
int n;
circles(){}
void add(circle cc){
c[n++] = cc;
}
//x 包含在 y 中
bool inner(circle x,circle y){
if(x.relationcircle(y) != 1)return 0;
return sgn(x.r-y.r)<=0?1:0;
}
//圆的面积并去掉内含的圆
void init_or(){
bool mark[maxn] = {0};
int i,j,k=0;
for(i = 0;i < n;i++){
for(j = 0;j < n;j++)
if(i != j && !mark[j]){
if( (c[i]==c[j])||inner(c[i],c[j]) )break;
}
if(j < n)mark[i] = 1;
}
for(i = 0;i < n;i++)
if(!mark[i])
c[k++] = c[i];
n = k;
}
//圆的面积交去掉内含的圆
void init_add(){
int i,j,k;
bool mark[maxn] = {0};
for(i = 0;i < n;i++){
for(j = 0;j < n;j++)
if(i != j && !mark[j]){
if( (c[i]==c[j])||inner(c[j],c[i]) )break;
}
if(j < n)mark[i] = 1;
}
for(i = 0;i < n;i++)
if(!mark[i])
c[k++] = c[i];
n = k;
}
//半径为 r 的圆,弧度为 th 对应的弓形的面积
double areaarc(double th,double r){
return 0.5*r*r*(th-sin(th));
}
//测试 SPOJVCIRCLES SPOJCIRUT
//SPOJVCIRCLES 求 n 个圆并的面积,需要加上 init_or() 去掉重复圆(否则WA)
//SPOJCIRUT 是求被覆盖 k 次的面积,不能加 init_or()
//对于求覆盖多少次面积的问题,不能解决相同圆,而且不能 init_or()
//求多圆面积并,需要 init_or, 其中一个目的就是去掉相同圆
void getarea(){
memset(ans,0,sizeof(ans));
vector<pair<double,int> >v;
for(int i = 0;i < n;i++){
v.clear();
v.push_back(make_pair(-pi,1));
v.push_back(make_pair(pi,-1));
for(int j = 0;j < n;j++)
if(i != j){
Point q = (c[j].p-c[i].p);
double ab = q.len(),ac = c[i].r, bc = c[j].r;
if(sgn(ab+ac-bc)<=0){
v.push_back(make_pair(-pi,1));
v.push_back(make_pair(pi,-1));
continue;
}
if(sgn(ab+bc-ac)<=0)continue;
if(sgn(ab-ac-bc)>0)continue;
double th = atan2(q.y,q.x), fai = acos((ac*ac+
ab*ab-bc*bc)/(2.0*ac*ab));
double a0 = th-fai;
if(sgn(a0+pi)<0)a0+=2*pi;
double a1 = th+fai;
if(sgn(a1-pi)>0)a1-=2*pi;
if(sgn(a0-a1)>0){
v.push_back(make_pair(a0,1));
v.push_back(make_pair(pi,-1));
v.push_back(make_pair(-pi,1));
v.push_back(make_pair(a1,-1));
}
else{
v.push_back(make_pair(a0,1));
v.push_back(make_pair(a1,-1));
}
}
sort(v.begin(),v.end());
int cur = 0;
for(int j = 0;j < v.size();j++){
if(cur && sgn(v[j].first-pre[cur])){
ans[cur] += areaarc(v[j].first-pre[cur],c[i].r)
;
ans[cur] += 0.5*(Point(c[i].p.x+c[i].r*cos(pre[
cur]),c[i].p.y+c[i].r*sin(pre[cur]))^Point(c
[i].p.x+c[i].r*cos(v[j].first),c[i].p.y+c[i
].r*sin(v[j].first)));
}
cur += v[j].second;
pre[cur] = v[j].first;
}
}
for(int i = 1;i < n;i++)
ans[i]-= ans[i+1];
}
};

//.3平面最近点对
const int MAXN = 100010;
const double INF = 1e20;
struct Point{
double x,y;
void input(){
scanf("%lf%lf",&x,&y);
}
};
double dist(Point a,Point b){
return sqrt((a.x-b.x)*(a.x-b.x) + (a.y-b.y)*(a.y-b.y));
}
Point p[MAXN];
Point tmpt[MAXN];
bool cmpx(Point a,Point b){
return a.x < b.x || (a.x == b.x && a.y < b.y);
}
bool cmpy(Point a,Point b){
return a.y < b.y || (a.y == b.y && a.x < b.x);
}
double Closest_Pair(int left,int right){
double d = INF;
if(left == right)return d;
if(left+1 == right)return dist(p[left],p[right]);
int mid = (left+right)/2;
double d1 = Closest_Pair(left,mid);
double d2 = Closest_Pair(mid+1,right);
d = min(d1,d2);
int cnt = 0;
for(int i = left;i <= right;i++){
if(fabs(p[mid].x-p[i].x) <= d)
tmpt[cnt++] = p[i];
}
sort(tmpt,tmpt+cnt,cmpy);
for(int i = 0;i < cnt;i++){
for(int j = i+1;j < cnt && tmpt[j].y-tmpt[i].y < d;j++)
d = min(d,dist(tmpt[i],tmpt[j]));
}
return d;
}
int main(){
int n;
while(scanf("%d",&n) == 1 && n){
for(int i = 0;i < n;i++)p[i].input();
sort(p,p+n,cmpx);
printf("%.2lf\n",Closest_Pair(0,n-1));
}
return 0;
}

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//三维几何
using namespace std;
const double eps = 1e-8;
int sgn(double x){
if(fabs(x) < eps)return 0;
if(x < 0)return-1;
else return 1;
}
struct Point3{
double x,y,z;
Point3(double _x = 0,double _y = 0,double _z = 0){
x = _x;
y = _y;
z = _z;
}
void input(){
scanf("%lf%lf%lf",&x,&y,&z);
}
void output(){
scanf("%.2lf-%.2lf-%.2lf\n",x,y,z);
}
bool operator ==(const Point3 &b)const{
return sgn(x-b.x) == 0 && sgn(y-b.y) == 0 && sgn(z-b.z) ==0;
}
bool operator <(const Point3 &b)const{
return sgn(x-b.x)==0?(sgn(y-b.y)==0?sgn(z-b.z)<0:y<b.y):x<b.x;
}
double len(){
return sqrt(x*x+y*y+z*z);
}
double len2(){
return x*x+y*y+z*z;
}
double distance(const Point3 &b)const{
return sqrt((x-b.x)*(x-b.x)+(y-b.y)*(y-b.y)+(z-b.z)*(z-b.z)
);
}
Point3 operator-(const Point3 &b)const{
return Point3(x-b.x,y-b.y,z-b.z);
}
Point3 operator +(const Point3 &b)const{
return Point3(x+b.x,y+b.y,z+b.z);
}
Point3 operator *(const double &k)const{
return Point3(x*k,y*k,z*k);
}
Point3 operator /(const double &k)const{
return Point3(x/k,y/k,z/k);
}
//点乘
double operator *(const Point3 &b)const{
return x*b.x+y*b.y+z*b.z;
}
//叉乘
Point3 operator ^(const Point3 &b)const{
return Point3(y*b.z-z*b.y,z*b.x-x*b.z,x*b.y-y*b.x);
}
double rad(Point3 a,Point3 b){
Point3 p = (*this);
return acos( ( (a-p)*(b-p) )/ (a.distance(p)*b.distance(p))
);
}
//变换长度
Point3 trunc(double r){
double l = len();
if(!sgn(l))return *this;
r /= l;
return Point3(x*r,y*r,z*r);
}
};
struct Line3
{
Point3 s,e;
Line3(){}
Line3(Point3 _s,Point3 _e)
{
s = _s;
e = _e;
}
bool operator ==(const Line3 v)
{
return (s==v.s)&&(e==v.e);
}
void input()
{
s.input();
e.input();
}
double length()
{
return s.distance(e);
}
//点到直线距离
double dispointtoline(Point3 p)
{
return ((e-s)^(p-s)).len()/s.distance(e);
}
//点到线段距离
double dispointtoseg(Point3 p)
{
if(sgn((p-s)*(e-s)) < 0 || sgn((p-e)*(s-e)) < 0)
return min(p.distance(s),e.distance(p));
return dispointtoline(p);
}
//返回点 p 在直线上的投影
Point3 lineprog(Point3 p)
{
return s + ( ((e-s)*((e-s)*(p-s)))/((e-s).len2()) );
}
//p 绕此向量逆时针 arg 角度
Point3 rotate(Point3 p,double ang)
{
if(sgn(((s-p)^(e-p)).len()) == 0)return p;
Point3 f1 = (e-s)^(p-s);
Point3 f2 = (e-s)^(f1);
double len = ((s-p)^(e-p)).len()/s.distance(e);
f1 = f1.trunc(len); f2 = f2.trunc(len);
Point3 h = p+f2;
Point3 pp = h+f1;
return h + ((p-h)*cos(ang)) + ((pp-h)*sin(ang));
}
//点在直线上
bool pointonseg(Point3 p)
{
return sgn( ((s-p)^(e-p)).len() ) == 0 && sgn((s-p)*(e-p))
== 0;
}
};
struct Plane
{
Point3 a,b,c,o;//平面上的三个点,以及法向量
Plane(){}
Plane(Point3 _a,Point3 _b,Point3 _c)
{
a = _a;
b = _b;
c = _c;
o = pvec();
}
Point3 pvec()
{
return (b-a)^(c-a);
}
//ax+by+cz+d = 0
Plane(double _a,double _b,double _c,double _d)
{
o = Point3(_a,_b,_c);
if(sgn(_a) != 0)
a = Point3((-_d-_c-_b)/_a,1,1);
else if(sgn(_b) != 0)
a = Point3(1,(-_d-_c-_a)/_b,1);
else if(sgn(_c) != 0)
a = Point3(1,1,(-_d-_a-_b)/_c);
}
//点在平面上的判断
bool pointonplane(Point3 p)
{
return sgn((p-a)*o) == 0;
}
//两平面夹角
double angleplane(Plane f)
{
return acos(o*f.o)/(o.len()*f.o.len());
}
//平面和直线的交点,返回值是交点个数
int crossline(Line3 u,Point3 &p)
{
double x = o*(u.e-a);
double y = o*(u.s-a);
double d = x-y;
if(sgn(d) == 0)return 0;
p = ((u.s*x)-(u.e*y))/d;
return 1;
}
//点到平面最近点 (也就是投影)
Point3 pointtoplane(Point3 p)
{
Line3 u = Line3(p,p+o);
crossline(u,p);
return p;
}
//平面和平面的交线
int crossplane(Plane f,Line3 &u)
{
Point3 oo = o^f.o;
Point3 v = o^oo;
double d = fabs(f.o*v);
if(sgn(d) == 0)return 0;
Point3 q = a + (v*(f.o*(f.a-a))/d);
u = Line3(q,q+oo);
return 1;
}
};